Let $Z\sim\mathcal{N}(0,1)$ (i.e. a random variable which distribution is the standard normal distribution). Determine the characteristical function of $Z$.
It is $\mathbb{P}_Z=f\lambda$ with $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$ and $\lambda$ the Lebesgue-mesaure. $$ \varphi_Z(t)=\int e^{itz}\, d\mathbb{P}_Z=\int e^{itz}f\, d\lambda $$ Now in a textbook I saw that they simply write $$ \int e^{itz}f\, d\lambda=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{itz}e^{-\frac{z^2}{2}}\, dz. $$
Why can they switch over to the Riemann-integral here?
And why integrating over $(-\infty,\infty)$? I do not see that $Z$ has to be a random variable on $\mathbb{R}$.
The last formula is nothing more than a tautology. Recall that $d\lambda$ is the Lebesgue measure on the line. Also, the integrand function is continuous, hence Riemann integrable. Therefore the Lebesgue integral reduces to a Riemann integral.